Dear student, 93 % (w/v), H2SO4 means 93 g of H2SO4 is present in 100 ml of solution Volume = 1000 ml Amount of H2SO4 = 930 g/ 1000 mlDensity = 1 84 g/mlM = density × volumeM = 1 84 × 1000 = 1840 gWeight of solvent = 1840 - 930 = 910 gMolality = Moles of solute Weight of solvent (in kg)Molality = 93098×1000910 = 10 43 mol/kg Regards

Help with this

Dear student,

$93 % (w / v), H_{2} S O_{4} m e a n s 93 g o f H_{2} S O_{4} i s p r e s e n t i n 100 m l o f s o l u t i o n . V o l u m e = 1000 m l A m o u n t o f H_{2} S O_{4} = 930 g / 1000 m l D e n s i t y = 1.84 g / m l M = d e n s i t y \times v o l u m e M = 1.84 \times 1000 = 1840 g W e i g h t o f s o l v e n t = 1840 - 930 = 910 g M o l a l i t y = \frac{M o l e s o f s o l u t e}{W e i g h t o f s o l v e n t (i n k g)} M o l a l i t y = \frac{930}{98} \times \frac{1000}{910} = 10.43 m o l / k g$
Regards

-1

View Full Answer