Chemistry, 25.03.2020 05:31 wallsdeandre25521

Hydrochloric acid is usually purchased in concentrated form with a 37.0% HCl concentration by mass and a density of 1.20 g / mL. How much of the concentrated stock solution in milliliters should you use to make 2.5 L of 0.500 M HCl

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Answer from: nett4386

The volume of the concentrated stock solution needed to prepare the solution is 102.71 mL

Let the mass of the concentrated HCl solution be 100 g.

thus, the mass of 37.0% HCl will be 37 g

Next, we shall determine the volume of the concentrated HCl solution

Mass of solution = 100 g

Density of solution = 1.20 g/ mL

Volume of solution = ?

$volume \: = \frac{mass}{density} \\ \\ volume \: = \frac{100}{1.2} \\ \\$

Volume of solution = 83.33 mLNext, we shall determine the number of mole in 37 g of HCl. This can be obtained as follow:

Mass of HCl = 37 g

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

mole of HCl =?

$mole \: = \frac{mass}{molar \: mass} \\ \\ mole \: = \frac{37}{36.5} \\ \\$

Mole of HCl = 1.014 moleNext, we shall determine the molarity of the concentrated HCl solution.

Mole of HCl = 1.014 mole

Volume of solution = 83.33 mL = 83.33 / 1000 = 0.08333 L

Molarity of HCl = ?

$molarity \: = \frac{mole}{volume} \\ \\ molarity \: = \frac{1.014}{0.08333}$

Molarity of HCl = 12.17 MFinally, we shall determine the volume of stock solution needed.

Molarity of stock solution (M₁) = 12.17 M

Volume of diluted solution (V₂) = 2.5 L = 2.5 × 1000 = 2500 mL

Molarity of diluted solution (M₂) = 0.5 M

Volume of stock solution needed (V₁) = ?

M₁V₁ = M₂V₂

12.17 × V₁ = 0.5 × 2500

12.17 × V₁ = 1250

Divide both side by 12.17

V₁ = 1250 / 12.17

V₁ = 102.71 mL

Therefore, the volume of the concentrated stock solution needed to prepare the solution is 102.71 mL

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Answer from: makailaaa2

0.10 L

Explanation:

Given: 37 gram of HCl are dissolved in 100 g of solution.

Density of solution= 1.20 g/ml

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.20g/ml}=83.3ml$

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\tims 1000}{V_s}$

where,

n = moles of solute = $\frac{\text{Given mass}}{\text{Molar mass}}=\frac{37g}{36.5}=1.01moles$

V_s = volume of solution in ml = 83.3 ml

$Molarity=\frac{1.01\tims 1000}{83.3}=12.2M$

M_1V_1(stock)=M_2V_2(dilute)

$12.2\times V_1=0.500\times 2.5L$

V_1=0.10L

0.10 L of concentrated stock solution in milliliters should you use to make 2.5 L of 0.500 M HCl

Answer from: Quest

answer: e. sour taste

explanation: acids are those substances which dissociates in water to give $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

$hcl\rightarrow h^++cl^-$

bases are those substances which dissociates in water to give $oh^-$ ions or accepts $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.