Calculate the approximate number of molecules in a drop of water with a mass of 0.10 g.

Chemistry

1 answer:

DENIUS [597]2 years ago

3 0

Answer:

The correct answer is B) 3 x 1021 molecules

Explanation:

We use Avogadro's number (in one mole of compound there are 6, 02x10e23 molecules) and calculate the weight of one mole of water:

Weight 1 mol H20 = (Weight H)x2 + Weight 0= 1 g/mol x 2 + 16 g/mol=

Weight 1 mol H20 = 18g/mol

18g---------6,02x10e23 molecules

0,10g-------x=(0,10gx6,02x10e23 molecules)/18g=3,34x10e21 molecules

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uranmaximum [27]

Answer:

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Explanation:

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Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$