A colourless salt X has 50% `Na_(2)SO_(3)` and 50% `H_(2)O` . How much of `SO_(2)` at NTP is obtained when 2.52 g of X reacts with excess of dil . `H_(2)SO_(4)` ?A. 22.4 LB. 0.448 LC. 44.8 LD. 0.224 L

A colourless salt X has 50% `Na_(2)SO_(3)` and 50% `H_(2)O` . How much

1.	A colourless salt X has 50% `Na_(2)SO_(3)` and 50% `H_(2)O` . How much of `SO_(2)` at NTP is obtained when 2.52 g of X reacts with excess of dil . `H_(2)SO_(4)` ?A. 22.4 LB. 0.448 LC. 44.8 LD. 0.224 L
Answer» Correct Answer - D In salt (X)=`underset(50%)(Na_(2)SO_(3)):underset(50%)(H_(2)O)` Molar mass of the `Na_(2)SO_(3)=126` Molar mass of the `H_(2)O=18` Hence , 7 molecules of water are attached with `Na_(2)SO_(3)` `therefore ` Formula =`Na_(2)SO_(3)7H_(2)O` `126:18xx 7 implies 126:126implies 1:1` `Na_(2)SO_(3)7H_(2)O + H_(2)SO_(4)(aq) to SO_(2)(aq) + Na_(2)SO_(3)(aq) + H_(2)O(l)` `because 252 g Na_(2)SO_(3)* 7H_(2)O ` gives `=22.4L SO_(2)` `therefore 2.52 g Na_(2)SO_(3) *7H_(2)O` will give `=(22.4)/(252)xx252=0.224 L`

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A colourless salt X has 50% `Na_(2)SO_(3)` and 50% `H_(2)O` . How much of `SO_(2)` at NTP is obtained when 2.52 g of X reacts with excess of dil . `H_(2)SO_(4)` ?A. 22.4 LB. 0.448 LC. 44.8 LD. 0.224 L

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