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4 L of 0.02 M aqueous solution of NaCl was diluted by adding 1 L of water. The molality of the resultant solution is……..A. 0.004B. 0.008C. 0.012D. 0.016 |
Answer» Correct Answer - D `M_1V_1=M_2V_2` 0.02 M x 4 L = `M_2 ` x 5 L `M_2=(0.02xx4)/5`=0.016 M Here, molality will be equal to molarity as for NaCl, molecular weight is equal to its equivalent weight. |
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