Question

Refer to the attachment and solve plz.​

Answer 1

Solution :

Firstly let's understand the reaction of the question

$\sf\:KOH+HCl\:\longrightarrow\:KCl+H_2O$

Now , Part -1 of the question

Molarity of of KOH = 0.2

it means ,No of moles of KOH of 1000 ml = 0.2

Thus , No of moles of KOH in 500 ml = 0.1

Similarly,

Molarity of of HCl = 0.2

it means ,No of moles of HCl of 1000 ml = 0.2

Thus , No of moles of HCl in 500 ml = 0.1

Now , Total volume of the solution = 1000 ml = 1 l

Weight of the solution = 1000 gm

Let the Heat of neutralisation in this case be $\sf\:T_1$

We know that,

Heat of neutralisation :

Q=mC∆t

$\sf\:Q=1000CT_1$

$\sf\:T_1=\dfrac{Q}{1000C}....(1)$

Now , part -2 of the question

Experiment is repeated by 250 ml each of ghr solution .

Molarity of of KOH = 0.2

it means ,No of moles of KOH of 1000ml = 0.2

Thus , No of moles of KOH in 250 ml = 0.05

Similarly,

Molarity of of HCl = 0.2

it means ,No of moles of HCl of 1000 ml = 0.2

Thus , No of moles of HCl in 250 ml = 0.05

Now , Total volume of the solution = 500 ml

Weight of the solution = 500 gm

Let the Heat of neutralisation in this case be $\sf\:T_2$

In this reaction , the moles of KOH and HCl in this solution are half to that of the 1st solution

Therefore, the heat of neutralisation will be half of the intital reaction thus ,

$\sf\:\dfrac{Q}{2}=500CT_2$

$\sf\implies\:T_2=\dfrac{Q}{1000C}...(2)$

Now divide Equation (1) & (2) , then

$\sf\:\dfrac{T_1}{T_2}=\dfrac{1}{1}$

Given

$\sf\:\dfrac{T_1}{T_2}=\dfrac{x}{y}$

On comparing x = 1

${\boxed{\sf{\red{therefore\:the\:value\:of\:x\:is\:1}}}}$