According to boiling point elevation equation:
Δ T = i Kb m
when ΔT (change in boiling point) = 7.10 C°
and i (van't Hoff factor)= 1
and Kb = 0.520
so, by substitution:
7.10 = 1*0.520 *m
m = 7.1 / 0.52 = 13.65 m

7 0

3 years ago

Determine the number of moles within a sample of 10g of propane(C3H8). Round your answer to 2 decimal places with proper units.

a_sh-v [17]

Given :

10 gram sample of propane( C₂H₈ ).

To Find :

The number of moles of propane in given sample.

Solution :

Molecular mass of propane, M = (2 × 12) + ( 1 × 8 ) gram/mol

M = 32 gram/mol

We know, number of moles is given by :

Number of moles, n = m/M

n = 10/32 mol

n = 0.3125 mol

Therefore, number of moles in given sample is 0.3125 mol.

7 0

2 years ago

At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO

erastovalidia [21]

Answer : The concentration of $NO$ at equilibrium is 0.9332 M

Solution : Given,

Concentration of $N_2$ and $O_2$ at equilibrium = 0.200 M

Concentration of $N_2$ and $O_2$ at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}$

$K_c=6.25$

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially 0.200 0.200 0

.800

At equilibrium (0.200-x) (0.200-x) (0.800+2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}$

By solving the term x, we get

$x=2.6\text{ and }-0.0666$

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of $NO$ at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of $NO$ at equilibrium is 0.9332 M

5 0

2 years ago