Answer:
i cant under stand what you are trying to say
Explanation:
Answer:
The correct answer is 1.977 % m/v ≅ 2% m/v
Explanation:
We have:
0.617 M = 0.617 moles methanol/ 1 L solution
We need:
%m/v= grams of methanol/100 mL solution
So, first we convert the moles of methanol to grams by using the MM (32.04 g/mol). Then, we multiply by 0,1 to convert the volume in liters to 100 mL by using the ratio: 100 mL= 0.1 L:
0.617 mol / 1 L x 32.04 g/mol 0.1 L/100 mL= 1.977 g/100 mL= %m/v
<span>Answer:
W must be 5-chloro-2-methylpentane. It can give only 4-methy-1-pentene (Y) upon dehydrohalogenation:
X must be 4-chloro-2-methylpentane. Dehydrohalogenation yields both Y and 4-methyl-2-pentene. (Z)</span>
We are going to use this equation:
ΔT = - i m Kf
when m is the molality of a solution
i = 2
and ΔT is the change in melting point = T2- 0 °C
and Kf is cryoscopic constant = 1.86C/m
now we need to calculate the molality so we have to get the moles of NaCl first:
moles of NaCl = mass / molar mass
= 3.5 g / 58.44
= 0.0599 moles
when the density of water = 1 g / mL and the volume =230 L
∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg
now we can get the molality = moles NaCl / Kg water
=0.0599moles/0.23Kg
= 0.26 m
∴T2-0 = - 2 * 0.26 *1.86
∴T2 = -0.967 °C
Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
