Question

Solve this Numerical From Solutions Class 12th..

Any one of these,You may Also Attempt Two.​

Answer 1

HEY THERE

2.) given- WB=60g , WA=250g , kf=1.86 , MB=108g

formula given-

$delta \: tf = \frac{kf \times wb \times100 }{mb \times wa}$

we get,

$deta \: tf = \frac{1.86kg \: per \: {mol}^{ - 1} \times 60g \times 1000g \: kg^{ - 1} }{108 g {mol}^{ - 1} \times 250g}$

$delta \: tf = 2583.33k$

freezing point of the solution,

tf= T knot f-delta Tf

=273.15-2583.33

=1310.18k

《HOPE IT HELP》

#i will edit my answer when i got another question #

^_^

Answer 2

HEY THERE

2.) given- WB=60g , WA=250g , kf=1.86 , MB=108g

formula given-

delta \: tf = \frac{kf \times wb \times100 }{mb \times wa}deltatf=

mb×wa

kf×wb×100

we get,

deta \: tf = \frac{1.86kg \: per \: {mol}^{ - 1} \times 60g \times 1000g \: kg^{ - 1} }{108 g {mol}^{ - 1} \times 250g}detatf=

108gmol

−1

×250g

1.86kgpermol

−1

×60g×1000gkg

−1

delta \: tf = 2583.33kdeltatf=2583.33k

freezing point of the solution,

tf= T knot f-delta Tf

mark me plz

=273.15-2583.33

=1310.18k