What is the final solution molarity when 75mL of isopropyl alcohol (C3H8O) (dens

Question

What is the final solution molarity when 75mL of isopropyl alcohol (C3H8O) (density of.785g/mol) is mixed with water to prepare a final solution volume of 1.0 Liters? I got 1.05 Molarity. Is this correct?

Explanation / Answer

First step is to find the molar mass of C3H8O by looking at the periodic table. Carbon(12.01) x 3(number of carbon molecules)+Hydrogen(1.08) x 8+Oxygen(16.00)=53.11g C3H8O Second step is to use dimensional analysis to convert grams------>moles and moles---->molecules and molecules---->atoms so, you take your given 72.5g of isopropanol and multiply it by 1 mole of C3H8O divided by your molar mass(53.11). That will cancel out grams and leave you with moles. Next, you use avogadro's number(6.022x10^23) and divide it by one mole wich will cancel out the moles and leave you with molecules. Lastly, you take the number of H atoms and divide by the one mole of C3H8O to get your final answer. 72.5g x 1mol C3H8O/53.11g C3H8O x 6.022x10^23molecules/1 mole x 8Hydrogen atoms/1molecule C3H8O =6.01x10^24

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