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What volume of 0.300M would contain 1.5g
(H=1, S=32, O=16​

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To calculate the volume of 0.300M solution that contains 1.5g of the solute, we need to use the following formula:

moles of solute = mass of solute / molar mass of solute

moles of solute = (1.5g) / (1 x H + 1 x S + 3 x O) g/mol

moles of solute = (1.5g) / (1 + 32 + 3x16) g/mol

moles of solute = (1.5g) / 98 g/mol

moles of solute = 0.015306 moles

Now, we can use the following formula to calculate the volume of the solution:

moles of solute = molarity x volume (in liters)

0.015306 moles = 0.300 M x volume (in liters)

volume (in liters) = 0.015306 moles / 0.300 M

volume (in liters) = 0.05102 L

Finally, we can convert the volume from liters to milliliters (mL):

volume (in mL) = 0.05102 L x 1000 mL/L

volume (in mL) = 51.02 mL

Therefore, 51.02 mL of 0.300M solution would contain 1.5g of solute.

User Tassos Bassoukos
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