The pair of optical Isomers which differ in the orientation of H and OH gp only at C1 Carbon atom are called anomers. Difference between two anomers of glucose :

Besides thymine, the two other products are : 2-deoxy-Dribose and phosphoric acid.

x₁ + x₂ + x₃ = 1

Sucrose < Glucose < Urea < NaCl

\(\pi =CRT=\frac { n }{ V } RT\)
Here, number of moles of solute dissolved (n)  \(=\frac { 1.0 \ g }{ 185,000 \ g \ { mol }^{ -1 } } =\frac { 1 }{ 185,000 } mol\)
\(V=450 \ mL=0.450 \ L,T={ 37 }^{ \circ }C=37+273=310 \ K\)
\( R=8.314 \ kPa \ L \ { K }^{ -1 }{ mol }^{ -1 }=8.314\times { 10 }^{ 3 }Pa \ L \ { K }^{ -1 }{ mol }^{ -1 }\)
Substituting these values, we get
\(\pi =\frac { 1 }{ 185,000 } mol \times\frac { 1 }{ 0.45 \ L }x \times8.314 \times { 10 }^{ 3 }Pa \ L \ { K }^{ -1 }{ mol }^{ -1 } \times 310 \ K=30.96 \ Pa.\)

Here,  \({ K }_{ H }=4.27\times { 10 }^{ 5 }mm,\ p=760 \ mm\)
Applying Henry's law, \(p={ K }_{ H }x, we \ have \ x=\frac { p }{ { K }_{ H } } =\frac { 760 \ mm }{ 4.27 \times{ 10 }^{ 5 }mm } =1.78 \times{ 10 }^{ -3 }\)
i.e., mole fraction of methane in benzene = 1.78 x 10^-3

A solution is a homogeneous mixture of two or more substances whose composition can be varied. On the basis of physical state of the component of solution, there are the following nine types of solutions:
(i) Gases Solutions 

(ii) Liquid Solutions

(ii) Solid Solutions

The cyclic hemiacetal form of glucose contains an OH group at C-I which gets hydrolysed in aqueous solution to produce the open chain aldehydic form which then reacts with \({ NH }_{ 2 }OH\) to form the

corresponding oxime. Thus, glucose contains an aldehydic group. In contrast, when glucose is reacted with acetic anhydride, the OH group at C-l, along with the four other OH groups at C-2, C-3, C-4 and C-6 form a pentaacetate. Since the pentaacetate of glucose does not contain a free OH group at C-l, it cannot get hydrolysed in aqueous solution to produce the open chain aldehydic form and hence glucose pentaacetate does not react with \({ NH }_{ 2 }OH\) to form glucose oxime.
This proves that glucose pentaacetate does not contain the aldehyde group

(a)(i) Colligative properties: Those properties of solutions which depend upon the number of particles of solute and solvent but not on the nature of solute are called colligative properties.
(ii) Molality of a solution: It is defined as the number of moles of solute per 1000 g or 1 kg of solvent \(m={n_B\over W_A}\times 1000\ \ or\ \ m={W_B\over M_B\times W_A}\times 1000\)
where 'n_B' is the number of moles of solute, W_A is the weight of solvent in grams, W_B is the amount of solute and 'M_B' is molecular weight of solute.
(b) P_N = 0.78 atm = 0.78 x 760 mm Hg
 = 592.8 mm Hg
K_H = 8.42 x 10^-7 M/mm Hg
X_N2 =?
X_N2 = K_H X P_N2
[Since KH is given in M/mm Hg, therefore, this formula is being used]
=> x_N = 8.42 x 10-7 M/mm Hg x 592.8mm Hg
=> X_N2 = 4991.376 x 10^-7 M
= 4.99 x 10^-4 M
Also, \(x_{N_2}={n_{N_2}\over n_{N_2}+n_{H_20}}\)
\(={n_{N_2}\over n_{H_2O}}={n_{N_2}\over {1000\over 18}}\)
\(n_{N_2}={1000\over 18}\times 4.99 \times 10^{-4}\)
= 0.0277 M = 2.77 x 10^-2 M

\(\text { Number of moles of acetic acid } =\frac{0.6 \mathrm{~mL} \times 1.06 \mathrm{~g} \mathrm{~mL}^{-1}}{60 \mathrm{~g} \mathrm{~mol}^{-1}} \\\)
\(=0.0106 \mathrm{~mol}=n\)
\(\text { Molality }=\frac{0.0106 \mathrm{~mol}}{1000 \mathrm{~mL} \times 1 \mathrm{~g} \mathrm{~mL}^{-1}}=0.0106 \mathrm{~mol} \mathrm{~kg}^{-1}\)
\(\triangle\)T_f = 1.86 K kg mol^–1 x 0.0106 mol kg^–1 = 0.0197 K
\(\text { van't Hoff Factor }(i)=\frac{\text { Observed freezing point }}{\text { Calculated freezing point }}=\frac{0.0205 \mathrm{~K}}{0.0197 \mathrm{~K}}=1.041\)
Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen ions per molecule of acetic acid. If x is the degree of dissociation of acetic acid, then we would have n (1 – x) moles of undissociated acetic acid, nx moles of CH₃COO^– and nx moles of H⁺ ions,

Thus total moles of particles are: n(1 – x + x + x) = n(1 + x)
\(i=\frac{n(1+x)}{n}=1+x=1.041\)
Thus degree of dissociation of acetic acid = x = 1.041– 1.000 = 0.041
Then [CH₃COOH] = n(1 – x) = 0.0106 (1 – 0.041),
[CH₃COO^–] = nx = 0.0106 x 0.041, [H⁺] = nx = 0.0106 x 0.041.
\(\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=\frac{0.0106 \times 0.041 \times 0.0106 \times 0.041}{0.0106(1.00-0.041)}\)
= 1.86 × 10^–5

Mass of solute, \({ \omega }_{ B }=5g\)
Mass of water, \({ \omega }_{ A }=100g\)
Molecular mass of water, \({ M }_{ A }=18\)
Vapour pressure of water, \({ p }_{ A }^{ 0 }=23.755\quad torr\)
Vapour pressure of solution, \({ p }_{ A }=23.402\quad torr\)
Lowering in vapour pressure = \({ p }_{ A }^{ 0 }-{ p }_{ A }=23.755 -23.402=0.353\)
\(Now,\frac { { p }_{ A }^{ 0 }-{ p }_{ A } }{ { p }_{ A }^{ 0 } } =\frac { { \omega }_{ B }{ M }_{ A } }{ { \omega }_{ A }{ M }_{ B } }\)
\( \frac { 0.353 }{ 23.755 } =\frac { 5\times 18 }{ 100\times { M }_{ B } }\)
\( \\ or \ { M }_{ B }=\frac { 5\times 18 }{ 100\times 0.353 } \times 23.755=60.56\)

The various quantities known to us are as follows: P = 2.57 × 10^–3 bar,
V = 200 cm³ = 0.200 litre
T = 300 K
R = 0.083 L bar mol^-1 K^-1
Substituting these values in equation (2.42) we get
\(M_2=\frac{1.26 \mathrm{~g} \times 0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{2.57 \times 10^{-3} \mathrm{bar} \times 0.200 \mathrm{~L}}=61,022 \mathrm{~g} \mathrm{~mol}^{-1}\)

(i) Peptide linkage. Peptide bond is formed by the condensation of two or more same or different n-amino acids. The condensation occurs between amino acids with the elimination of water. In this case, the carboxyl group of one amino acid and amino group of another amino acid get condensed with the elimination of water molecule.
The resulting \(\quad O\\ \quad \parallel \\ -C-NH-\)  linkage is called peptide linkage. The formation of a dipeptide and the peptide 

(ii) Primary structure. The primary structure of proteins gives the sequence in which the amino acids are linked in one or more polypeptide chains of proteins. This is shown below:

(iii) Denaturation. A process that changes the physical and biological properties of proteins without affecting the chemical composition of a protein is called denaturation. The denaturation is caused by certain physical or chemical treatments such as changes in pH, temperature, presence of some salts or certain chemical agents.

0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH₂CONH₂) = 60 g mol^-1
\(\therefore\) 0.25 mole of urea = 0.25 x 60 = 15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
\(\therefore\)2.5 kg of solution contains urea = 15/1.015 x 2.5 = 37 g

(i) Carrot and Cod liver oil.
(ii) Nucleotides are monomers of nucleic acids. They consist of heterocyclic base, pentose sugar and phosphoric acid residue.

(ii) Glucose on oxidation with HNO₃ gives saccharic acid. This indicates the presence of a one primary alcoholic (-OH) group in glucose.
(iii) The pentaacetate of glucose does not react with hydroxylamine which shows the absence of free aldehydic (-CHO) group.
(iv) In \(\alpha\)-D-glucose, the -OH group at C₁ is towards right, whereas in \(\beta\)-D-glucose, the -OH group at C₁ is towards left. Such a pair of stereoisomers which differ in the configuration only at C₁ are called anomers.

Or
Actually, glucose exists in the cyclic hemiacetal form (hence the aldehyde group is not free) with only a small amount (<0.05%) of the open chain form.
Since, the concentration of the open chain form is low and its reaction with 2,4-DNP is reversible, therefore, formation of 2,4-DNP derivative cannot disturb the equilibrium to regenerate more open chain form from the cyclic hemiacetal form and hence, does not give this test.