Asked by Kevin
Calculate the molarity of 3 drops (0.15mL) of a 0.5M Na3PO4 solution after adding 15 drops (.75mL) of water.
(Presume drops are the same size, one drop = 0.05mL)((Presume volumes are additive))
So far I've figured .15g Na3PO4/163.94g/mol Na3PO4 = .0009moles
.75g H20/18.015 = 0.0416 mol H20
What are the next steps to solve this problem? Thank you!
(Presume drops are the same size, one drop = 0.05mL)((Presume volumes are additive))
So far I've figured .15g Na3PO4/163.94g/mol Na3PO4 = .0009moles
.75g H20/18.015 = 0.0416 mol H20
What are the next steps to solve this problem? Thank you!
Answers
Answered by
bobpursley
you are diluting the sodium phosphate by a factor of .90/.15=6
that means, the new molarity is .5/6
that means, the new molarity is .5/6
Answered by
Kevin
Thanks Bob!
Curious where .90 comes from? .0009molX1000mL?
.5/6= .083M
Curious where .90 comes from? .0009molX1000mL?
.5/6= .083M
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