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2 years ago

A sample of helium has a volume of 3.49 L at 741 mmHg. What would be the volume if the pressure

Chemistry

1 answer:

Naddika [18.5K]2 years ago

5 0

P1: 741 mmHg

V1: 3.49 L P1 x V1 / P2= (741 mmHg) (3.49 L) / 760 mmHg = 3.40 L

P2: 760 mmHg

V2: ? L

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$\boxed{\text{3.3 mL}}$

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2. Moles of NaOH

$\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}$

3. Molar concentration of NaOH

$c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}$

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

$c_{1}V_{1} = c_{2}V_{2}$

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1 mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

$V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}$

(b) Calculate the volume of dilute solution

$\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}$

$\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}$

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