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Greeley [361]
3 years ago
14

What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4

.90 L of an HCl solution with a pH of 1.50? Express your answer to two significant figures and include the appropriate units.
Chemistry
1 answer:
Ivenika [448]3 years ago
5 0

Answer:

130ml of HCl(36%) in 4.90L solution => pH = 1.50

Explanation:

Need 4.90L of HCl(aq) solution with pH = 1.5.

Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺

[HCl(36%)] ≅ 12M in HCl

(M·V)concentrate = (M·V)diluted

12M·V(conc) = 0.032M·4.91L

=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.

Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.

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How many moles of molecules are in 10 g of aspartame?
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Answer:

molecules

Explanation:

Data Given:

mass of aspartame = 10 g

molecules of aspartame = ?

Solution

First we calculate no. of moles of 10 g aspartame

For which mole formula will be used

               no. of moles = mass in grams / molar mass . . . . . . .(1)

Formula of aspartame:

C₁₄H₁₈N₂O₅

So,

The molar mass of aspartame (C₁₄H₁₈N₂O₅) will be

molar mass of C₁₄H₁₈N₂O₅ = 14(12) + 18(1) + 2(14) +5(16)

molar mass of C₁₄H₁₈N₂O₅ = 168 + 18+ 28 + 80

molar mass of C₁₄H₁₈N₂O₅ = 294 g/mol

put values in equation 1

          no. of moles = 10 g / 294 g/mol

          no. of moles = 0.034 mol

now we will calculate no. of molecules

Formula will be used

           no. of moles = no. of molecules / Avogadro's number

Rearrange the above equation:

no. of molecules = no. of moles x Avogadro's number . . . . . (2)

Where

Avogadro's number = 6.022 x 10²³

Put values in equation 2

           no. of molecules =0.034 mol x 6.022 x 10²³ (molecules/mol)

           no. of molecules = 2.0475 x 10²²

So,

There are 2.0475 x 10²² molecules are in 10 grams of aspartame that is 0.034 moles of aspartame.

So,  in turn its 0.034 moles of molecules are in 10 g of aspartame.

8 0
3 years ago
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