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### A 6.50 molal solution of KOH (aq) has density of $1.89 g.cm^{-3}$ ....

A 6.50 molal solution of KOH (aq) has density of $1.89 g.cm^{-3}$. The molarity of 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]

Solution

We have, $6.50 = \frac{{w \times 1000}}{{56 \times W}}$

$\therefore \frac{W}{w} = \frac{{1000}}{{6.50 \times 56}}$

Density $= \frac{{w + W}}{V} = 1.89 \times 1000$ g/ltr

Molarity $= \frac{w}{{56 \times V}} = \frac{{w \times 1.89 \times 1000}}{{56 \times (w + W)}} = \frac{{1.89 \times 1000}}{{56 \times \left( {1 + \frac{W}{w}} \right)}} = \frac{{1.89 \times 1000}}{{56 \times \left( {1 + \frac{{1000}}{{6.50 \times 56}}} \right)}} = 9$

[Note: $ltr = dm^3 = 1000 cm^3$]

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$