What would be the concentration of salt if $174 \cdot g$ $174g$ $N a C l$ $NaCl$ were added to $1000 \cdot m L$ $1000mL$ of water?

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What would be the concentration of salt if $174 \cdot g$ $174g$ $N a C l$ $NaCl$ were added to $1000 \cdot m L$ $1000mL$ of water?

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Answer 1 · 2016-03-01T16:37:26

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $("Moles of solute")/("Volume of solution")$

Explanation:

If we have $1.00$ $1.00$ $k g$ $kg$ of water, there is $1.00$ $1.00$ $L$ $L$ of water. (Note that volume would not change substantially upon dissolution of the salt).

Thus, $Molarity$ $"Molarity"$ $=$ $=$ $((174*cancelg)/(58.44*cancelg*mol^-1))/(1.00*L)$ $~~$ $3$ $mol*L^-1$

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What would be the concentration of salt if $174 \cdot g$ $174g$ $N a C l$ $NaCl$ were added to $1000 \cdot m L$ $1000mL$ of water?

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What would be the concentration of salt if 174*g174⋅g174*g NaClNaClNaCl were added to 1000*mL1000⋅mL1000*mL of water?

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What would be the concentration of salt if $174 \cdot g$ $174g$ $N a C l$ $NaCl$ were added to $1000 \cdot m L$ $1000mL$ of water?