Question

the crystalline salt na2so4.xH2o on heating lose 55.9% of its mass and becomes anhydrous.the formula of salt is ??

Answer 1

Molar mass of salt = (2 × 23) + 32 + (16 × 4) + x × 18
= 142 + 18x
On heating it will lose water and become anhydrous. 55.9% mass is the mass of water in Na2SO4.xH2O.
Mass by mass% = (mass / molar mass of compound) × 100
mass of water = 18x
55.9 = (18x / 142 + 18x) × 100
55.9(142 + 18x) = 1800x
7937.8 + 1006.2x = 1800x
x = 9.99
x = 10 (approx.)
So, the molecular formula of compound is Na2SO4.10H2O.

Answer 2

The formula of salt is $\bold{Na_2SO_4.1H_2O}$

Given:

The molecular formula of crystalline salt = $Na_2SO_4.xH_2O$

The lose weight of salt on heating = 55.9 %

Solution:

Let's calculate the molar mass of $Na_2SO_4.xH_2O$

$Molar\quad mass =(2 \times 23)+32+(16 \times 4)+\mathrm{x} \times 18$

$Molar\quad mass =142+18 \mathrm{x}$

By heating of salt becomes anhydrous form.

$Mass\quad by\quad mass\quad percentage\quad =\quad \left( \frac { { Mass } }{ { Molar\quad mass\quad of\quad compound } } \right) \quad \times \quad 100$

$Mass\quad of\quad water\quad =\quad 18{ x }$

$55.9=\left(\frac{18 x}{142+18 x}\right) \times 100$

$55.9(142+18 x)=1800 x$

$7937.8+1006.2 \mathrm{x}=1800 \mathrm{x}$

$793.7=1800 x-1006.2 x$

$793.7=793.8 x$

$x=\frac{793.7}{793.8}$

$x=0.99 \approx 1$

From the calculation x value is 1, substitute the 'x' value in salt formula.

Therefore, the salt formula will be $\bold{Na_2SO_4.1H_2O}$