1. the molecular formula of sulphuric acid is H2SO4: (a) calculate the molecular mass of sulphuric acid. (b) how many moles are there in 9.8g of sulphuric acid ? (c) how many "gram-atoms" of each element are there in 9.8g of sulphuric acid ? (atomic masses: H = 1, S = 32, O = 16) 2. Find the number of molecules present in a drop of chloroform (CHCl3) weighing 0.0239g. (C = 12, Cl = 35.5, H = 1)

_{2}SO

_{4}) = 2$\times $1 + 32 + 16$\times $4 = 98 u

b) Given mass of H

_{2}SO

_{4}= 9.8 g

Number of moles (n) of H

_{2}SO

_{4}= $\frac{\text{Givenmass}}{\text{Molarmass}}$

n = $\frac{9.8}{98}$ = 0.1 moles

c) 98 g of H

_{2}SO

_{4}contains = 32 g of sulphur

9.8 g of H

_{2}SO

_{4}contains = $\frac{32}{98}\times 9.8$ = 3.1 gram sulphur atoms

98 g of H

_{2}SO

_{4}

_{ }contains = 2 g of hydrogen atoms

9.8 g of H

_{2}SO

_{4}contains = $\frac{2}{98}\times 9.8$ = 0.2 gram hydrogen atoms

98 g of H

_{2}SO

_{4}contains = 64 g of oxygen atoms

9.8 g of H

_{2}SO

_{4}contains = $\frac{64}{98}\times 9.8$ = 6.39 gram oxygen atoms

2) Molar mass of chloroform (CHCl

_{3}) = 12 + 1 + 3$\times $35.5 = 119.5 g mol

^{-1}

119.5 g of CHCl

_{3}contains = 6.022$\times $10

^{23}molecules

0.0239 g of CHCl

_{3}= $\frac{6.022\times {10}^{23}}{119.5}\times 0.0239$ = 0.0012 molecules

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