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7nadin3 [17]
2 years ago
6

A student takes a stock solution that is 50 mM (solute formula weight is 120.5 g/mol) and prepares a series of solutions. The fi

rst solution is made by diluting 1 mL of the stock in water to a final volume of 15 mL (sample 1). They then take a 2 mL aliquot of sample 1 and dilute in water to a final volume of 25 mL (sample 2). Finally, they take a 1.5 mL aliquot of sample 2 and dilute to a final volume of 250 mL. Calculate the final molar concentration for the analyte in molarity, molality, ppm, and ppb. State any assumptions.
Chemistry
1 answer:
Ira Lisetskai [31]2 years ago
3 0

Answer:

1.6x10⁻⁶M

1.6x10⁻⁶m

0.1928ppm

192.8ppb

Explanation:

The first dilution of the solution is from 1mL to 15mL. The second from 2mL to 25mL and the third from 1.5mL to 250mL. That means molarity is:

50mM =

0.050M * (1mL / 15mL) * (2mL / 25mL) * (1.5mL / 250mL) = 1.6x10⁻⁶M

Molality is defined as the ratio between moles and kg. Assuming the density of the solution is 1kg/L, molality is 1.6x10⁻⁶m

ppm is the ratio between milligrams and Liters:

1.6x10⁻⁶mol / L * (120.5g /mol) * (1000mg / g) = 0.1928mg/L = 0.1928ppm

And ppb = 1000*ppm;

0.1928ppm*1000 = 192.8ppb

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WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
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How many electrons are in their outer shell?
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Is there an attachment because it depends on the element.
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Hexane and octane are mixed to form a 45 mol% hexane solution at 25 deg C. The densities of hexane and octane are 0.655 g/cm3 an
avanturin [10]

Answer:

The required volume of hexane is 0.66245 Liters.

Explanation:

Volume of octane = v=1.0 L=1000 cm^3

Density of octane= d = 0.703 g/cm^3

Mass of octane ,m= d\times v=0.703 g/cm^3\times 1000 cm^3=703 g

Moles of octane =\frac{m}{114 g/mol}=\frac{703 g}{114 g/mol}=6.166 mol

Mole percentage of Hexane = 45%

Mole percentage of octane = 100% - 45% = 55%

55\%=\frac{6.166 mol}{\text{Total moles}}\times 100

Total moles = 11.212 mol

Moles of hexane :

45%=\frac{\text{moles of hexane }}{\text{Total moles}}\times 100

Moles of hexane = 5.0454 mol

Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g

Density of the hexane,D = 0.655 g/cm^3

Volume of hexane = V

V=\frac{M}{D}=\frac{433.9044 g}{0.655 g/cm^3}=662.4494 cm^3\approx 0.66245 L

(1 cm^3= 0.001 L)

The required volume of hexane is 0.66245 Liters.

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