You dissolve 4.5 mol of potassium hydroxide to make a 2 L solution. What is the molarity of the solution?

1 answer:

7 0

Data:
M (molarity) = ? (Mol/L)
m (mass) = ? (in grams)
V (volume) = 2 L
MM (Molar Mass) of Potassium Hydroxide (KOH)
K = 1*39 = 39 amu
O = 1*16 = 16 amu
H = 1*1 = 1 amu
------------------------------------
MM of KOH = 39+16+1 = 56 g/mol

***Let's find the mass
Data:
n (number of mols) = 4.5 mols
m (mass) = ? (in grams)
v (volume) = 2 L

Formula:

$n = \frac{m}{V}$
$4.5 = \frac{m}{2}$
$m = 4.5*2$
$\boxed{m = 9.0\:g}$

***Let's find the molarity:

Formula:

$M = \frac{m}{MM*V}$

Solving:
$M = \frac{m}{MM*V}$
$M = \frac{9.0}{56*2}$
$M = \frac{9.0}{112}$
$M = 0.0803...\:\to\:\boxed{\boxed{M \approx 0.08\:Mol/L}}\end{array}}\qquad\quad\checkmark$

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Calculate the rate of dissolution (dM/dt) of relatively hydrophobic drug particles with a surface area of 2.5×103 cm2 and satura

Crank

Answer:

$\large \boxed{\text{1.22 mg/s}}$

Explanation:

We can use the Noyes-Whitney equation to calculate the rate of dissolution.

$\dfrac{\text{d}M}{\text{d}t} = \dfrac{DA(C_{s} - C)}{d}$

Data:

D = 1.75 × 10⁻⁷ cm²s⁻¹

A = 2.5 × 10³ cm²

Cₛ = 0.35 mg/mL

C = 2.1 × 10⁻⁴ mg/mL

d = 1.25 µm

Calculations:

Cₛ - C = (0.35 - 2.1 × 10⁻⁴) mg·cm⁻³ = 0.350 mg·cm⁻³

d = 1.25 µm = 1.25 × 10⁻⁶ m = 1.25 × 10⁻⁴ cm

$\dfrac{\text{d}M}{\text{d}t} = \dfrac{(1.75 \times 10^{-7} \text{cm}^{2}\text{s}^{-1})(2.5 \times 10^{3} \text{ cm}^{2})(0.350\text{ mg$\cdot$cm$^{-3}$})}{1.25 \times 10^{-4} \text{ cm}} = \textbf{1.22 mg/s}\\\\\text{The rate of dissolution is $\large \boxed{\textbf{1.22 mg/s}}$}$