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`25.3 g` sodium carbonate, `Na_(2)CO_(3)`, was dissolved in enough water to make `250 mL` of solution. If sodium carbonate dissociates completely, molar concentration of `Na^(+)` and carbonate ions are respectively:A. 0.477 M and 0.477 MB. 0.955 M and 1.910 MC. 1.910 M and 0.955 MD. 1.90 M and 1.910 M |
Answer» Correct Answer - 3 `M(Na_(2)CO_(3))=n_(Na_(2)CO_(3))/V_(mL)xx(1000 mL)/L` `=(25.3 g Na_(2)CO_(3)//106 g Na_(2)CO_(3) mol^(-1))/(250 mL)xx(1000 mL)/L` `=0.955 mol L^(-1) Na_(2)CO_(3)(aq) rarr 2Na^(+)(aq)+CO_(3)^(2-)(aq)` `1 mol 2 mol 1 mol` Thus `M(Na^(+))=2M(Na_(2)CO_(3))=2(0.955 M)=1.910 M` `M(CO_(3)^(2-))=M(Na_(2)CO_(3))` `=0.955 M` |
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