The hydrated salt $N a_{2} S O_{4}, 10 H_{2} O$ undergoes $X %$ loss in weight on heating and becomes anhydrous. The value of $X$ will be:

10

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45

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56

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70

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Solution

The correct option is A $56$
Molecular mass of $N a_{2} S O_{4} 10 H_{2} O = 2 \times 23 + 32 + 16 \times 14 + 1 \times 20 = 322$

On heating the $N a_{2} S O_{4} 10 H_{2} O$ it becomes anhydrous that means 10 $H_{2} O$ molecules separated

Molecular mass of $10 H_{2} O = 20 \times 1 + 10 \times 16 = 180$

% loss $= \frac{180}{322} \times 100 = 55.9$ % $= 56$ %

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The hydrated salt Na2SO4,10H2O undergoes X% loss in weight on heating and becomes anhydrous. The value of X will be:

The correct option is A 56Molecular mass of Na2SO410H2O=2×23+32+16×14+1×20=322On heating the Na2SO410H2O it becomes anhydrous that means 10 H2O molecules separatedMolecular mass of 10H2O=20×1+10×16=180% loss =180322×100=55.9 % =56 %

The hydrated salt $N a_{2} S O_{4}, 10 H_{2} O$ undergoes $X %$ loss in weight on heating and becomes anhydrous. The value of $X$ will be: