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JEE Main 202429 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2024 · 29 January · Evening Shift · Chemistry · Solutions

Molality of 0 . 8M H _ 2 SO _ 4 solution (density 1 . 06g cm ^ - 3...

Molality of 0.8MH2SO40 . 8M{H}_{2}{SO}_{4} solution (density 1.06gcm−31 . 06g{cm}^{- 3} ) is _______×10−3m\times {10}^{- 3}m.

Round off your answer to the nearest integer.

Correct answer

0

Step-by-step solution

10001000 ml solution contains 0.80 . 8mol of H2SO4

Mass of solution = 100cm3×1.06gcm3=1060g100{cm}^{3} \times 1 . 06\dfrac{g}{{cm}^{3}} = 1060g

Mass of solute (H2SO4) =0.8×98=78.4g0 . 8 \times 98 = 78 . 4g

Mass of solvent = 1060−78.4=981.6g1060 - 78 . 4 = 981 . 6g

Molality = \dfrac{molesof{H}_{2}{SO}_{4}}{massofsolvent ( inkg )}</p><p></p><p>= \dfrac{0 . 8}{981 . 6 \times {10}^{- 3}} = 0 . 815m</p><p>=</p><p>=815 \times {10}^{- 3}m

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