how do I do this problem
0.700 mol of Na2S in 1.55 of solution?
if you want M, then just recall that M = mol/L
.7mol/1.55L = 0.45 mol/L = 0.45M
To solve this problem, you need to use the given information and perform a few calculations. Here's a step-by-step guide:
1. First, convert the given amount of sodium sulfide (Na2S) from moles to grams. To do this, you need to know the molar mass of Na2S, which consists of two sodium (Na) atoms with a molar mass of 22.99 g/mol each, and one sulfur (S) atom with a molar mass of 32.07 g/mol.
Molar mass of Na2S = (2 * molar mass of Na) + molar mass of S
= (2 * 22.99 g/mol) + 32.07 g/mol
= 45.98 g/mol + 32.07 g/mol
= 78.05 g/mol
Now, you can calculate the mass of 0.700 moles of Na2S using the following equation:
Mass = Number of moles * Molar mass
= 0.700 mol * 78.05 g/mol
= 54.64 g (rounded to two decimal places)
Therefore, 0.700 moles of Na2S is equal to 54.64 grams.
2. Next, find the concentration of the solution in terms of grams per liter (g/L). The concentration can be calculated by dividing the mass of the solute (Na2S) by the volume of the solution (1.55 L).
Concentration = Mass of Solute / Volume of Solution
= 54.64 g / 1.55 L
= 35.35 g/L (rounded to two decimal places)
Therefore, the concentration of the solution is 35.35 g/L.
To summarize, to solve the problem you need to convert the given moles of Na2S to grams, and then divide that mass by the volume of the solution to find the concentration in grams per liter.