If the boiling point of water is 100 ,^{circ} | vedclass

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(D) The elevation in boiling point is given by the formula: $\Delta T_{b} = i 	imes K_{b} 	imes m$. For $NaCl$,the van't Hoff factor $i = 2$. The mo

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$28.12$

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(D) The elevation in boiling point is given by the formula: $\Delta T_{b} = i 	imes K_{b} 	imes m$. For $NaCl$,the van't Hoff factor $i = 2$. The molality $m$ is defined as: $m = \frac{W_{NaCl} 	imes 1000}{M_{NaCl} 	imes W_{H_2O(g)}}$. Given $\Delta T_{b} = 1 \,K$,$K_{b} = 0.52 \,K \cdot kg/mol$,$M_{NaCl} = 58.5 \,g/mol$,and $W_{H_2O} = 500 \,g$. Substituting the values: $1 = 2 	imes 0.52 	imes \frac{W_{NaCl} 	imes 1000}{58.5 	imes 500}$. Solving for $W_{NaCl}$: $W_{NaCl} = \frac{1 	imes 58.5 	imes 500}{2 	imes 0.52 	imes 1000} = 28.125 \,g$.

If the boiling point of water is $100 \,^{\circ}C$,how many grams of $NaCl$ must be added to $500 \,g$ of water to increase the boiling point by approximately $1 \,^{\circ}C$ (in $,g$)? Given $(K_{b})_{H_2O} = 0.52 \,K \cdot kg/mol$.

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