This is due to the presence of dust particles.

1 mile/h = \(\frac { 1.6\times { 10 }^{ 3 }m }{ 3600s } \) = 0.444ms^-1
100 mile/h = 44.4 m/s

The compound is glucose.Its molecular formula is C₆H₁₂O₆ while empirical formula is CH₂O.

1 mole of carbon atoms = 6.023 × 10²³ atoms of carbon
Mass of 1 atom of 12C =Atomic mass of C/Avogadro′s number
\(=  \frac{ 12}{6.022×10^{23}} g\)
= 1.9927 ×10^-23 g

In case of solution,molarity is calculated by using molarity equation M₁V₁ = M₂V₂, we have V₁(before dilution) and V₂ (after dilution) so calculate molarity of the given solution from this equation.
Given that, M₁ = 5 M \(\Rightarrow \) V₁ = 500 mL
V₂ = 1500 mL \(\Rightarrow \) M₂ = M
For dilution, a general formula is
\(\underset { Before \ diution }{ { M }_{ 1 }{ V }_{ 1 }= } \underset { \quad After \ diution }{ { M }_{ 2 }{ V }_{ 2 } } \) 
\(500\times 5M=1500\times M\Rightarrow M=\frac { 5 }{ 3 } =1.66M\)

1 carat = 200 mg, 
∴ 5 carat = 1000 mg = 1g
Hence, he should subtract 1 g from the weight of the ring to get the weight of gold.

Fe = 28% ; s = 24%

14 g

2.000 x 10³

8.721 x 10⁴

Concentration of sugar
Given, mass of salute =20 g
Molarmassofsolute(C₂H₂₂O₁₁)=12×12+22×1+11×16=342 g/mol
Volume of solution (inL)=2
We know,
Molarity(M)\(=\frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}\)
\(=\frac{\frac{\text{Mass of sugar }}{\text{Molar mass of sugar}}}{2L}\)
\(=\frac{\frac{20 g}{342 g}}{2L}\)
=0.02925 molL⁻¹
Final answer: Molar concentration =0.02925 mol L^–1

0.375M solution means 0.375 moles in 1 L of solution.
1000mL of solution contains 0.375 moles
\(1 \mathrm{~mL} \text { of solution } \longrightarrow \frac{0.375}{1000} \text { moles }\)
\(500 \mathrm{~mL} \text { of solution } \longrightarrow \frac{0.375}{1000} \times 500\)
⟶ 0.1875
\(\text { No. of moles }=\frac{\text { Mass }}{\text { Molar Mass }}\)
\(0.1875 \mathrm{~mol}=\frac{\text { Mass }}{82.0245 \mathrm{gmol}^{-1}}\)
∴ Mass of the compound = 15.37 g.

a) Cost of table salt (NaCl) per mole
Gram molecular mass of NaCl = 23 + 35.5 = 58.5 g
Now, 1000 g of NaCl cost = Rs.2
\(\therefore \) 58.5 g of NaCl will cost = \(\frac { 2 }{ \left( 1000 \ g \right) } \times \left( 58.5 \ g \right) \) = 0.117
= 0.117 x 100 = 12 paise (approx.)
(b) Cost of table sugar (C₁₂H₂₂O₁₁) per mole
Gram molecular mass of (C₁₂H₂₂O11) = 12 x 12 + 22 x 1 = 16 x 11
= 144 + 22 + 176 = 342 g
Now, 1000 g of sugar cost = Rs.6
\(\therefore \) 342 g of sugar will cost =\(\frac { 6 }{ \left( 1000 \ g \right) } \times \left( 342 \ g \right) \)=2.052
= 2.0 Rupees (approx.)