All star marked Questions are multi...

All star marked Questions are multiple correct with one or more than one Answer. 0.1 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?

JEE/Engineering Exams
4.0 (1 ratings)
Mslaily = '/ by meight X density x 10 [ begin{array}{l} =frac{13 times 1.02 times 10}{98} =frac{132.6}{98}=1.35 mathrm{M} end{array} ] 13% solution by weight means anat isg of solute is exter dissolucd un ( 87 mathrm{g} ) for soluent Nemaliliy ( 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 cdot 70 N=1.5 mathrm{N} times V_{2} sigma_{2}=180 mathrm{mL} end{array} ] 6 &o the aid should be diluted vpto is ML tr Mepare 1.5 N soln
Quick and Stepwise Solutions Just click and Send Download App OVER 20 LAKH QUESTIONS ANSWERED Download App for Free