Question

50ml of 0.1N solution contain 0.31g of Na2CO3.xH2O.Find the molecular mass and value of x​

Answer 1

Given:

50ml of 0.1N solution contain 0.31g of Na2CO3.xH2O.

To find:

Value of x

Calculation:

Number of equivalents involved

$= \dfrac{50}{1000} \times 0.1$

$= 0.05 \times 0.1$

$= 5 \times {10}^{ - 3} \: eqv.$

Now, we know that:

$\rm{ \therefore \: \dfrac{given \: mass}{equivalent \: mass} = no. \: of \: equivalents}$

$\rm{ = > \: \dfrac{0.31}{equivalent \: mass} = 5 \times {10}^{ - 3} }$

$\rm{ = > \:equivalent \: mass = \dfrac{0.31}{5 \times {10}^{ - 3} } }$

$\rm{ = > \:equivalent \: mass =62}$

Now , equivalent mass of anhydrous Na2CO3 is

$\rm{ \therefore \: equivalent \: mass = \dfrac{106}{2} }$

$\rm{ = > \: equivalent \: mass = 53}$

So, value of x

$\boxed{ \bf{x = 62 - 53 = 9}}$