A laboratory procedure calls for making 400.0 mL of a 1.1 M
NaNO3...

Question

A laboratory procedure calls for making 400.0 mL of a 1.1 M
NaNO3 solution. What mass of NaNO3 (in g) is needed?

Accepted Answer

Hi everyone. So ask calculate the mass in grams. A cesium sulfate. That is needed for a pair of 15 ml of 56 smaller cesium sulfate solution. Recall that malarkey. You go to malls of solute. I thought about leaders of solution. Since we're giving the volume in milliliters need to convert from milliliters to leaders 12 15 male leaders. And in 1000 ml We have one leader To give us 0.015 leaders. And then we have the volume in leaders and the polarity. Who would use this assault for the moles of the solute? It's gonna be 56 smaller. That's gonna be equal to X. We're solving for the number of malls. About about 0.015 leaders. The most part 56 Time 0. leaders going to get X. An act Gonna be equal to 0.84 malls. So now that we have moles assessments of fate, we can convert to grams. Obsessing. So Faith You have 0. malls assessing sell fake And in one mold assessing sulfate their their roller mass. Which we need to find the Mueller mass. Gonna be equal to two. We have two sessions Times the mass a session which is 132 .905 grams Plus the massive sulfur. Which is 32 .066 grams Plus the mass of oxygen and therefore oxygen. So we're going to do four about 15 . gramps. No more masks. Number 361.8 g obsession. Your faith. And this will give us 303 . gramps. Thanks for watching my video, and I hope it was helpful.

A laboratory procedure calls for making 400.0 mL of a 1.1 M NaNO3 solution. What mass of NaNO3 (in g) is needed?

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