Question

the concentration of solution of sulphuric acid is 18M and has density 1.84gcm-3:what is the mole fraction and weight percentage of h2so4 in the solution​

Answer 1

We need to find the mole fraction of Suphuric Acid and it's Weight% . The concenⁿ of the Solution is 18M , in terms of Molarity which means , 18 moles of Suphuric Acid is present in one litre of solution. Hence , let's find out the weight of the solute dissolved. Using the formula of number of moles ,

$\sf\dashrightarrow n =\dfrac{Given\ wt.}{Molecular\ wt.} \\\\\sf\dashrightarrow 18 =\dfrac{W}{98g} \\\\\sf\dashrightarrow W = 98g\times 18 \\\\\sf\dashrightarrow \boxed{\sf Weight_{(given)}= 1764 \ grams }$

$\rule{200}2$

• Again it's given that , the density of the Solution is 1.84 g/cm³ . And we know that , density is mass divided by Volume. By this we can find the mass of the Solution. Here the Volume of Solution is 1L .

$\sf\dashrightarrow \rho =\dfrac{Mass}{Volume} \\\\\sf\dashrightarrow 1.84 \ g/cm^3= \dfrac{Mass}{1000 \ ml } \\\\\sf\dashrightarrow 1.84 \ g/cm^3= \dfrac{Mass}{1000 \ cm^3 } \\\\\sf\dashrightarrow Mass = 1.84 \times 1000 g \\\\\sf\dashrightarrow \boxed{\sf Mass_{(Solution)}= 1840g }$

Hence here we got the mass of Solution as 1840g and the mass of solute as 1746 g . So , the mass of solvent will be , 1840 g - 1746g = 76g.

$\rule{200}2$

Here the water is solvent. So the number of moles will be , 76/18 = 4.22 .

• And we know that , Mole Fraction is the ratio of no. of mole of the solvent to the no. of mole of the solution. So , that ,

$\sf\dashrightarrow \chi_{\tiny H_2SO_4}= \dfrac{n_{(H_2SO_4)}}{n_{(Solution)}} \\\\\sf\dashrightarrow \chi_{\tiny H_2SO_4}= \dfrac{ 18}{18+4.22} \\\\\sf\dashrightarrow \chi_{\tiny H_2SO_4}= \dfrac{ 18}{22.22} \\\\\sf\dashrightarrow \underset{\blue{\sf \tiny Required\ Answer}}{\underbrace{\boxed{\pink{\sf \chi_{\tiny H_2SO_4}= 0.81 }}}}$

$\rule{200}2$