if volume ratio of solute and solvent in solution is 4:5.Find v/v%
Answers
Answer:
80%
Explanation:
Volume Percent = Volume of Solute/Volume of Solution×100%
= 4/5×100
=80%
Answer:
Explanation:
Many people have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration of a solution is the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for solution reactions. Chemists use many different methods to define concentrations, some of which are described in this section.
Molarity
The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) is defined as the number of moles of solute present in exactly 1 L of solution. It is, equivalently, the number of millimoles of solute present in exactly 1 mL of solution:
\[ molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \label{4.5.1}\]
The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as \(M\). An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the molar concentration of a solute. Therefore,
\[[\rm{sucrose}] = 1.00\: M\]
is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either
\[ V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \label{4.5.2}\]
or
\[ V_{mL} M_{mmol/mL} = \cancel{mL} \left( \dfrac{mmol} {\cancel{mL}} \right) = mmoles \label{4.5.3}\]
Figure \(\PageIndex{1}\) illustrates the use of Equations \(\ref{4.5.2}\) and \(\ref{4.5.3}\).
alt
Figure \(\PageIndex{1}\): Preparation of a Solution of Known Concentration Using a Solid Solute
Example \(\PageIndex{1}\): Calculating Moles from Concentration of NaOH
Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.
Given: identity of solute and volume and molarity of solution
Asked for: amount of solute in moles
Strategy:
Use either Equation \ref{4.5.2} or Equation \ref{4.5.3}, depending on the units given in the problem.
Solution:
Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation \ref{4.5.2} is more useful:
\( moles\: NaOH = V_L M_{mol/L} = (2 .50\: \cancel{L} ) \left( \dfrac{0.100\: mol } {\cancel{L}} \right) = 0 .250\: mol\: NaOH \)
Exercise \(\PageIndex{1}\): Calculating Moles from Concentration of Alanine
Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.
Answer
Concentrations are also often reported on a mass-to-mass (m/m) basis or on a mass-to-volume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in parts per million (ppm), which is grams of solute per 106 g of solution, or in parts per billion (ppb), which is grams of solute per 109 g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per milliliter. These concentrations and their units are