Verified Solution • 2025–26 Session moderate

QJEEM21Chbff7: JEE Main 2021 Chemistry - Class 11 JEEM21Chbff7 Solution: A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm 3 …

A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm −- 3. The molarity of the solution is ____________ mol dm −- 3. (Round off to the Nearest Integer).

[Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u]

Step-by-Step Verified Solution

m=1000×M1000×d−M×Msolutem = {{1000 \times M} \over {1000 \times d - M \times {M_{solute}}}}

6.5=1000×M1890−M×566.5 = {{1000 \times M} \over {1890 - M \times 56}}

⇒\Rightarrow 12285−364M=1000M12285 - 364M = 1000M

⇒\Rightarrow 1364M=122851364M = 12285

⇒\Rightarrow M=9M = 9

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