Question

On complete reaction with water, 0.1 g of KH gave a solution requiring $25c{m^3}$ of 0.1M HCl for neutralisation. Calculate the relative atomic mass of potassium from this information.
(A) 39
(B) 40
(C) 41
(D) None of these

Answer 1

Hint: First deduce the reaction involved in this question. Using that equation find the number of moles of the product that will be formed to neutralise 0.1 M HCl using the dilution formula of molarity. This will lead you to your answer.
  Complete answer: -Number of moles: Moles = given weight / molecular weight
-Dilution formula for molarity: ${M_1}{V_1} = {M_2}{V_2}$
  Where, M= molarity and V= Volume.

Complete step by step answer:
-First let us see what happens when KOH reacts with water:
                                    $KH + {H_2}O \to KOH + {H_2}$
-Let us assume the molar mass of K to be ‘x’. The weight of KH is given in the question to be 0.1 g. So, the molecular weight of KH will be= (M.wt. of K + M.wt. of H)
                                                                     = (x + 1) g
Then the number of moles of KH will be:
                               Moles of KH = given weight of KH / Molecular weight of KH
                                                     = 0.1 / (x + 1)
-Since according to the reaction 1 mole of KH gives 1 mole of KOH, then:
         [0.1 / (x + 1)] moles of KH will form [ 0.1 / (x + 1)] moles of KOH. (1)
The question says that the formed KOH requires 25 $c{m^3}$ of 0.1M HCl.
Thus, using the formula of dilution: ${M_1}{V_1} = {M_2}{V_2}$

     Where, ${M_1}$ = Molarity of HCl = 0.1 M
                    ${V_1}$ = Volume of HCl = 25 $c{m^3}$
                   ${M_2}$ = Molarity of KOH
                   ${V_2}$ = Volume of KOH
                     0.1×25 = ${M_2}{V_2}$
                   ${M_2}{V_2}$ = 2.5

[From the formula of molarity we can say that MV = n (moles)]
So, ${M_2}{V_2}$ is equal to the moles of KOH.
                             ${M_2}{V_2}$ = 2.5 millimoles of KOH
                              = 2.5 × ${10^{ - 3}}$ moles of KOH (2)
(millimoles because volume was in $c{m^3}$ which is equal to ml)
-We have 2 equations for moles of KOH, so let’s equate both these.
                      (1) = (2)
                                                     0.1 / (x + 1) = 2.5 × ${10^{ - 3}}$
   On cross multiplication: 0.1 / 2.5 × ${10^{ - 3}}$= x + 1
                                                   100 / 2.5 = x + 1
                                                              40 = x + 1
                                            So, x = 39
Since in the beginning we had assumed the molecular mass of K to be x. So, now we can say that the molecular mass of K is 39.
The correct option is: (A) 39

Note: Always pay attention to the units. In this question volume is given in ml.
Also the formula: ${M_1}{V_1} = {M_2}{V_2}$ , is used only when there is complete neutralisation of HCl by KOH and not otherwise.