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2 years ago

When 45.0g CaCl2 is dissolved to make a 750.mL solution what is the resulting molarity. Show work

1 answer:

katrin2010 [14]2 years ago

6 0

Molarity is the method of expressing concentration of solution. Mathematically, it is expressed as
Molarity = $\frac{\test{weight of solute (g)}}{\text{molecular weight X Volume of solution(l)}}$

In present case, Molecular weight of CaCl2 = 110.98 g
volume of solution = 750 ml = 0.750 l
weight of solute (CaCl2) = 45 g

∴Molarity = $\frac{\test{45}}{\text{110.98 X 0.750}}$
= 0.54 M

Concentration of solution is 0.54 M

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2 years ago

You have two compounds that you have spotted on the TLC plate. One compound is more polar than the other. You ran the TLC plate

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Answer:

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Explanation:

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2 years ago

A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight

zhannawk [14.2K]

Answer: The molecular formula will be $C_8H_8O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.9}{1.5}=4$

For H = $\frac{5.9}{1.5}=4$

For O = $\frac{1.5}{1.5}=1$

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is $C_4H_4O$

The empirical weight of $C_4H_4O$ = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2$

The molecular formula will be= $2\times C_4H_4O=C_8H_8O_2$

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2 years ago

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2 years ago

Read 2 more answers

How many molecules of CaCl2 are equivalent to 75.9g CaCl2 (Ca=40.08g/mol, CL=35.45g/mol)

melomori [17]

Answer:

$\large \boxed{4.12 \times 10^{23}\text{ formula unis of CaCl}_{2}}$}$

Explanation:

You must calculate the moles of CaCl₂, then convert to formula units of CaCl₂.

1. Molar mass of CaCl₂

CaCl₂ = 40.08 + 2×35.45 = 40.08 + 70.90 = 110.98 g/mol

2. Moles of CaCl₂ $\text{Moles of CaCl}_{2} = \text{75.9 g CaCl}_{2} \times \dfrac{\text{1 mol CaCl}_{2}}{\text{110.98 g CaCl}_{2}} = \text{0.6839 mol CaCl}_{2}$

3. Formula units of CaCl₂

$\text{No. of formula units} = \text{0.6839 mol CaCl}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules CaCl}_{2}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{4.12 \times 10^{23}}\textbf{ formula units CaCl}_{2}\\\text{There are $\large \boxed{\mathbf{4.12 \times 10^{23}}\textbf{ formula units of CaCl}_{2}}$}$

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2 years ago