jee-main 2025 Q55

jee-main · India · session2_02apr_shift2 Not Maths
Q55. 2.7 kg of each of water and acetic acid are mixed. The freezing point of the solution will be $- x ^ { \circ } \mathrm { C }$. Consider the acetic acid does not dimerise in water, nor dissociates in water. $x =$ $\_\_\_\_$ (nearest integer) [Given: Molar mass of water $= 18 \mathrm {~g} \mathrm {~mol} ^ { - 1 }$, acetic acid $= 60 \mathrm {~g} \mathrm {~mol} ^ { - 1 } \mathrm {~K} _ { \mathrm { f } } \mathrm { H } _ { 2 } \mathrm { O } : 1.86 \mathrm {~K} \mathrm {~kg} \mathrm {~mol} ^ { - 1 } \mathrm {~K} _ { \mathrm { f } }$ acetic acid: $3.90 \mathrm {~K} \mathrm {~kg} \mathrm {~mol} ^ { - 1 }$ freezing point: $\mathrm { H } _ { 2 } \mathrm { O } = 273 \mathrm {~K}$, acetic acid $= 290 \mathrm {~K}$ ]
Q55. 2.7 kg of each of water and acetic acid are mixed. The freezing point of the solution will be $- x ^ { \circ } \mathrm { C }$. Consider the acetic acid does not dimerise in water, nor dissociates in water. $x =$ $\_\_\_\_$ (nearest integer) [Given: Molar mass of water $= 18 \mathrm {~g} \mathrm {~mol} ^ { - 1 }$, acetic acid $= 60 \mathrm {~g} \mathrm {~mol} ^ { - 1 } \mathrm {~K} _ { \mathrm { f } } \mathrm { H } _ { 2 } \mathrm { O } : 1.86 \mathrm {~K} \mathrm {~kg} \mathrm {~mol} ^ { - 1 } \mathrm {~K} _ { \mathrm { f } }$ acetic acid:\\
$3.90 \mathrm {~K} \mathrm {~kg} \mathrm {~mol} ^ { - 1 }$ freezing point: $\mathrm { H } _ { 2 } \mathrm { O } = 273 \mathrm {~K}$, acetic acid $= 290 \mathrm {~K}$ ]\\