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A test solution contains 30 mg/cm^3 of Li+ ion. What mass of Li2SO4 is needed to make 750 cm^3 ?​

asked
User Griselda
by
8.2k points

1 Answer

1 vote

Final answer:

To calculate the mass of Li2SO4 needed to make 750 cm³ of a Li+ ion solution, we need to determine the number of moles of Li+ ions in the solution. From the given information, the concentration of Li+ ions is 30 mg/cm³. Using dimensional analysis, we can calculate the number of moles of Li+ ions in the solution.

Step-by-step explanation:

To calculate the mass of Li2SO4 needed to make 750 cm³ of a Li+ ion solution, we need to determine the number of moles of Li+ ions in the solution. From the given information, the concentration of Li+ ions is 30 mg/cm³. To convert this to moles, we need to know the molar mass of Li+.

Since Li+ is formed from Li, the molar mass of Li is 6.94 g/mol. Therefore, the molar mass of Li+ is also 6.94 g/mol. Using dimensional analysis, we can calculate the number of moles of Li+ ions in the solution as follows: Moles of Li+ = (Concentration of Li+ ions * Volume of solution) / Molar mass of Li+

answered
User Mark Loyman
by
7.6k points
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