JEE Main 202116 MarchMorning Shift

Exam: JEE Main
Year: 2021
Date: 16 March
Shift: Morning Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2021 · 16 March · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

A 6 . 50 molal solution of KO H(aq .) has a density of 1 . 89g cm ^...

A $6 . 50$ molal solution of $KO$ H\left(aq .\right) $has a density of$ 1 . 89g{cm}^{- 3} $. The molarity of the solution is$ mol{dm}^{- 3} $. (Round off to the Nearest Integer).</p><p>[Atomic masses:$ K: 39 . 0u; O: 16 . 0u; H: 1 . 0u]

Correct answer

Step-by-step solution

$6 . 5$ molal $KOH = 1000gm$ solvent has $6 . 5$ moles $KOH$

so, wt. of solute $= 6 . 5 \times 56$

$= 364gm$

wt. of solution $= 1000 + 364 = 1364$

Volume of solution $= \dfrac{1364}{1 . 89}mℓ$

Molarity $= \dfrac{moleof solute}{{V}_{solution}in Litre}$

$= 9 . 00$

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