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2 years ago

How many moles of solute particles are present in 1 mL (exact) of aqueous 0.0040 M Ba(OH)2?

Chemistry

1 answer:

lidiya [134]2 years ago

6 0

The number of mole of solute particles are present in 1 mL (exact) of aqueous 0.0040 M Ba(OH)₂ is 0.000004 mole

<h3>What is molarity? </h3>

Molarity is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

With the above formula, we can determine the number of mole present in the solution. Detail below:

<h3>How to determine the mole of the solute in the solution</h3>

Volume of solution = 1 mL = 1 / 1000 = 0.001 L
Molarity of solution = 0.004 M
Mole of solute =?

Molarity = mole / Volume

Cross multiply

Mole = Molarity × volume

Mole of solute = 0.004 × 0.001

Mole of solute = 0.000004 mole

Thus, the mole of the solute in the solution is 0.000004 mole

Learn more about molarity:

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A Geographic Information System is a system of which of these? A. Globes and maps B. Satellites C. Satellite signals and receive

AleksAgata [21]

Answer:

its D

Explanation:

the answer is D

7 0

4 years ago

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

3 0

3 years ago

Carbon is considered which of the following?

Zarrin [17]

Answer: atomic element

4 0

2 years ago

What is the mole ratio of PbO2 to water?<br><br> PbO, H,0 = 1:0

ipn [44]

Answer:

<em>The mole ratio of PbO2 to H2O is 1 : 2.</em>

Explanation:

The balanced reaction equation is:

Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O

On the reactant side, we have 1 mole of Pb, 1 mole of PbO2, 2 moles of H2SO4.

On the product side, we have 2 moles of PbSO4 and 2 moles of H2O.

This means that for ever 1 mole of PbO2 consumed, 2 moles of water is formed as product.

Hence, the mole ratio of PbO2 to H2O is 1 : 2.

5 0

3 years ago

Calculate the volume in dm3 occupied by 26.4 g of carbon dioxide

Mama L [17]

Density of carbon dioxide=1564kg/m^3
Mass=26.4g=0.0264kg

$\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}$

$\\ \sf\longmapsto Volume=\dfrac{Mass}{Density}$

$\\ \sf\longmapsto Volume=\dfrac{0.0264}{1564}$

$\\ \sf\longmapsto Volume=1.687m^3=1.69m^3$

1m^3=1000dm^3

$\\ \sf\longmapsto Volume=1.69(1000)=1690dm^3$

7 0

3 years ago