Part A Calculate the number of grams of solute in 0.290 L of 0.205 M K B r . m =

Question

Part A           Calculate  the number of grams of solute in 0.290L  of 0.205 M KBr.          m =
g
Part B           Calculate  the molar concentration of a solution containing 14.85g  of Ca(NO3)2 in 1.365L .         M =
M
Part C            Calculate the volume of 1.90 M Na3PO4 in milliliters that contains 4.00g  of solute.          V =
mL Part A           Calculate  the number of grams of solute in 0.290L  of 0.205 M KBr.          m =
g
Part A           Calculate  the number of grams of solute in 0.290L  of 0.205 M KBr.          m =
g
m =
g m =
g
Part B           Calculate  the molar concentration of a solution containing 14.85g  of Ca(NO3)2 in 1.365L .         M =
M
Part B           Calculate  the molar concentration of a solution containing 14.85g  of Ca(NO3)2 in 1.365L .         M =
M
M =
M M =
M
Part C            Calculate the volume of 1.90 M Na3PO4 in milliliters that contains 4.00g  of solute.          m =
g

Explanation / Answer

Molarity=moles of solute/volume of solution

moles=M x V

=0.205 x 0.290

=0.06

moles=weight/Mol.Weight

wieght=moles x M.W

=0.06 x 119

weight=7.07 g

M.W of Ca(N03)2=164 g

given weight =14.85 g

moles=weight/M.W

= 14.85/164

moles=0.09

given volume=1.365 L

Molarity=moles/volume

=0.09/1.365

Molarity=0.067 M

M.W of Na3P04=164 g

Given weight= 4 g

Molarity=1.90 M

MOlarity=molesx1000/volume(ml)

volume=moles x 1000/molarity

=weight x 1000/(molarityx M.w)

=4000/(1.90x164)

Volume =12.84 ml

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