The molar mass of a substance can be calculated by adding the atomic masses of all the atoms present in the atom.
1. H atomic mass =1
2. (S) atomic mass =32
3. O atomic mass =16
\(\mathrm{H}_2 \mathrm{SO}_4\) molar mass can be calculated by adding atomic masses of Hydrogen, Sulfur, and Oxygen = 2 x mass of H atom + mass of S atom +4 x mass of O atom =2 \times 1+32+4 \times 16=98
Hence, the molar mass of \(\mathrm{H}_2 \mathrm{SO}_4\) is 98 g.

(a) Steps:
108   Ã⋅7.2 = 14.5833
Three S.F. Two S.F.
Here answer should have two significant figures. Therefore, correct answer, after rounding off two significant figures is 15.
(b) Steps: 2 x 1.6 x 10²
2 x 163.2 = 326.4 = 300
(c) Steps: (1.0042 - 0.0034) x 1.23
1.0008 x 1.23 = 1.230984
= 1.23

(9.87 x 10^-3 - 2.26 x 10^-4)
= 9.87 x 10^-3 - 0.226 x 10^-3 = (9.87 - 0.226) x 10^-3
= 9.644 x 10^-3

(4.56 x 10³ + 2.62 x 10²)
= 45.6 x 10² + 2.62 x 10² = (45.6 + 2.62) x 10²
= 48.22 x 10² = 4.822 x 10³

The basic units for length is meter (m), for time is second(s) and for mass is kilogram (kg). 
\(25365mg\times \frac { 1g }{ 1000mg } \times \frac { 1kg }{ 1000g } =2.5365\times { 10 }^{ -2 }kg\)

There are three common scales to measure temperature ^oC (degree celsius), ^oF(degree Fahrenheit) and K (kelvin). The K is the SI unit.
The temperature on two scales are related to each other by the following relationship \(^{ 0 }F=\frac { 9 }{ 5 } t^{ 0 }C+32\) 
Putting the values in above equation.
\(200-32=\frac { 9 }{ 5 } t^{ 0 }C\Rightarrow \frac { 9 }{ 5 } t^{ 0 }C=168\)   
\(\Rightarrow \ t^{ 0 }C=\frac { 168\times 5 }{ 9 } =93.3^{ 0 }C\)

Suppose both read the same value as x.
Then as \(^{ 0 }C=\frac { 5 }{ 9 } (^{ 0 }F-32)\)
\(\therefore \)  x = 5/9 ( x - 32 ) or 9x = 5x - 160
or  4x = - 160 or x = -40^o.

OK, assuming it is 93% by mass then we should work with 100 g of solution a 93% by mass solution will have 93 g of H₂SO₄ and 7 g of H₂O
moles H₂SO₄ = mass / molar mass = 93 g / 98.086 g/mol = 0.948148 moles
mass H₂O = 7 g = 0.007 kg
molality = moles solute / kg solvent
= 0.948148 mol / 0.007 kg
= 135 m ~ 140 m (2 sig figs)
Now
total volume of 100 g of solution = mass / density
= 100 g / 1.84 g/ml
= 54.35 ml
= 0.05435 L
molarity = moles solute / litres solution
= 0.98148 mol / 0.05435 L
= 18 M