What is the value of freezing point depression of a solution of 15.0g of NaCl in 250g of water?

-3.82ºC is the freezing point of solution

Explanation:

We work with the Freezing point depression to solve the problem

ΔT = m . Kf . i

ΔT = Freezing point of pure solvent - freezing point of solution

Let's find out m, molality (moles of solute in 1kg of solvent)

15 g / 58.45 g/mol = 0.257 moles of NaCl

NaCl(s) → Na⁺ (aq) + Cl⁻(aq)

i = 2 (Van't Hoff factor, numbers of ions dissolved)

m = mol /kg → 0.257 mol / 0.250kg = 1.03 m

Kf = Cryoscopic constant → 1.86 ºC/m (pure, for water)

0ºC - Tºf = 1.03m . 1.86ºC/m . 2

Tºf  = -3.82ºC

linda: 3 ; laura: 15

step-by-step explanation:

note that linda + laura's age is 18, and that linda is 12 years younger than her sister.

let "linda" = x , "laura" = y:

x + y = 18

x = y - 12

first, plug in y - 12 for x in the first equation:

(y - 12) + y = 18

simplify.

2y - 12 = 18

isolate the variable (y). note the equal sign, what you do to one side, you do to the other. do the opposite of pemdas. first add, then divide.

2y - 12 (+12) = 18 (+12)

2y = 18 + 12

2y = 30

(2y)/2 = (30)/2

y = 30/2

y = 15

next, plug in 15 for y in the second equation (the other equation would work too).

x = y - 12

x = (15) - 12

x = 3

linda's age is 3, laura's age is 15.

~

m=9

step-by-step explanation: