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all star marked Questions are multiple correct with one or more than one Answer. What is the molarity and molality of a 13% solution (by weight) of sulphuric acid with a density of 1.02 g/ml To what volume should 100 ml of this acid be diluted in order to prepare a 1.5 N solution? [IIT-1978) ✓

JEE/Engineering Exams
Chemistry
Solution
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Mslaily = '/. by meyht X density x lo [ begin{array}{l} =frac{13 times 1.02 times 10}{98} =frac{132 cdot 6}{98}=1.35 mathrm{M} end{array} ] ( 13 % ) solution by weight means asat is a of solute is mat dissomed in ( 87 mathrm{g} ) for soluct Nemalily ( x ) eq. wit = molurity ( x ) wol. wet ( N times frac{98}{2}=1.35 times 98 ) [ begin{array}{c} N=1.35 times 2=2.70 mathrm{N} N_{1} V_{1}=N_{2} V_{2} 100 times 2.70 N=1.5 mathrm{N} times V_{2} V_{2}=180 mathrm{mL} end{array} ] Go the alid should be diluted upto is 80 ML tr Mepare 1.5 N soln
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