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Question: On complete reaction with water, 0.1 g of KH gave a solution requiring \(25c{m^3}\) of 0.1M HCl for ...

On complete reaction with water, 0.1 g of KH gave a solution requiring 25cm325c{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

Explanation

Solution

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: M1V1=M2V2{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+H2O→KOH+H2KH + {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 cm3c{m^3} of 0.1M HCl.
Thus, using the formula of dilution: M1V1=M2V2{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, M2V2{M_2}{V_2} is equal to the moles of KOH.
M2V2{M_2}{V_2} = 2.5 millimoles of KOH
= 2.5 × 10−3{10^{ - 3}} moles of KOH (2)
(millimoles because volume was in cm3c{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{10^{ - 3}}
On cross multiplication: 0.1 / 2.5 × 10−3{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: M1V1=M2V2{M_1}{V_1} = {M_2}{V_2} , is used only when there is complete neutralisation of HCl by KOH and not otherwise.