Solve C₁V₁ = C₂V₂ for any dilution variable, get molarity, % w/v and ppm from a weighed mass, the mass to weigh for a solution, or molality.
ਅੱਪਡੇਟ ਕੀਤਾ ਜਾਂਚੀਆਂ ਉਦਾਹਰਨਾਂ: 6
Take 25 mL of the 1 mol/L stock and make it up to 250 mL (add 225 mL of solvent) to get 0.1 mol/L, a 10× dilution.
Any consistent concentration unit works because it cancels; the working here is in mol/L and litres.
Strictly: make up to V₂ in a volumetric flask; volumes aren't always additive.
Diluting a solution leaves the moles of solute unchanged, so C₁V₁ = C₂V₂: stock concentration times stock volume equals final concentration times final volume. Enter any three and the fourth follows, with the solvent to add and the dilution factor C₁/C₂. The other modes turn a weighed mass into molarity (n = m/M, c = n/V), % w/v and mg/L, give the mass to weigh for a target molarity, or find molality per kilogram of solvent.
The default makes 250 mL of 0.1 mol/L from a 1 mol/L stock: measure 25 mL of stock and make it up to 250 mL, a tenfold dilution.
Volumes of mixed liquids do not always add exactly, so make up to the final volume in a volumetric flask rather than adding the calculated solvent volume. The ppm figure is mg per litre of solution, which equals mg/kg only for dilute water-based solutions.
ਜਾਂਚ ਦਾ ਸਰੋਤ: Python 3.8 decimal: V₁ = 0.1 × 250 / 1 = 25 mL (OpenStax Chem 2e §3.3 method)
ਜਾਂਚ ਦਾ ਸਰੋਤ: Python 3.8 decimal: 12 × 0.010 / 1
ਜਾਂਚ ਦਾ ਸਰੋਤ: C₁ = C₂ ⇒ V₁ = V₂
ਜਾਂਚ ਦਾ ਸਰੋਤ: Python 3.8 decimal: 5.844/58.440 g/mol/0.1 L = 1 M (NaCl = 22.990 + 35.45)
Write down the three values you know, use the same concentration unit on both sides and the same volume unit on both sides, and rearrange for the unknown. For 250 mL of 0.1 M from a 1 M stock, V₁ = C₂V₂ ÷ C₁ = 0.1 × 250 ÷ 1 = 25 mL of stock, made up to 250 mL with 225 mL of solvent. Because the units cancel, mmol/L or % work too.
Molarity (mol/L) is moles of solute per litre of solution; molality (mol/kg) is moles of solute per kilogram of solvent. Molality does not change with temperature because mass does not expand, which is why freezing-point and boiling-point calculations use it. For dilute water solutions the two are close: 58.44 g of NaCl in 2 kg of water is 0.5 mol/kg.
Multiply the target molarity by the volume in litres to get moles, then by the molar mass: m = c × V × M. For 500 mL of 0.2 M glucose (180.156 g/mol), 0.2 × 0.5 × 180.156 = 18.016 g. Dissolve it in less solvent than the final volume, then make up to the mark so the total is exactly 500 mL.
Multiply the molarity by the molar mass to get grams per litre, then by 1000 to get mg/L, which is ppm for dilute water solutions. A 0.001 M NaCl solution is 0.001 × 58.44 = 0.05844 g/L, or 58.44 ppm. Reading mg/L as ppm assumes a density of about 1 g/mL, so the two drift apart for concentrated or non-aqueous solutions.
It is the stock concentration divided by the final concentration, C₁/C₂, which equals the final volume divided by the stock volume, V₂/V₁. Making 25 mL up to 250 mL is a dilution factor of 10, often written 1:10, meaning 1 part stock in 10 parts total. Some labs write the same dilution as 1:9, meaning 1 part stock to 9 parts diluent, so check which convention a protocol uses.
ਸਟੀਕਤਾ ਤੁਹਾਡੇ ਇਨਪੁੱਟਾਂ ਅਤੇ ਵਿਧੀ ਦੀਆਂ ਧਾਰਨਾਵਾਂ ਉੱਤੇ ਨਿਰਭਰ ਕਰਦੀ ਹੈ। ਦਸ਼ਮਲਵ ਗਣਨਾ 50 ਮਹੱਤਵਪੂਰਨ ਅੰਕ ਵਰਤਦੀ ਹੈ, ਪਰ ਅਨੁਮਾਨ, ਅੰਕੀ ਵਿਧੀਆਂ ਅਤੇ ਸਰੋਤ ਡਾਟਾ ਘੱਟ ਸਟੀਕ ਹੋ ਸਕਦੇ ਹਨ; ਦਿਖਾਏ ਮੁੱਲਾਂ ਨੂੰ ਗੋਲ ਕਰਨ ਨਾਲ ਇਹ ਹੱਦਾਂ ਦੂਰ ਨਹੀਂ ਹੁੰਦੀਆਂ। ਸੁਤੰਤਰ ਸਰੋਤਾਂ ਦੀਆਂ ਹੱਲ ਕੀਤੀਆਂ ਉਦਾਹਰਨਾਂ ਨਾਲ ਜਾਂਚ: 6। ਉਦਾਹਰਨ ਲਈ, “250 mL of 0.1 M from a 1 M stock” ਦੀ ਜਾਂਚ Python 3.8 decimal: V₁ = 0.1 × 250 / 1 = 25 mL (OpenStax Chem 2e §3.3 method) ਨਾਲ ਕੀਤੀ ਜਾਂਦੀ ਹੈ।
OpenStax Chemistry 2e, §3.3 Molarity (dilution of solutions) and §3.4 Other units for solution concentrations; IUPAC Gold Book — amount concentration; molality.
ਇਸ ਕੈਲਕੁਲੇਟਰ ਵਿੱਚ ਸੁਤੰਤਰ ਸਰੋਤਾਂ ਦੇ ਜਵਾਬਾਂ ਵਾਲੀਆਂ 6 ਹੱਲ ਕੀਤੀਆਂ ਉਦਾਹਰਨਾਂ ਹਨ। ਇਹ ਟੈਸਟ ਸਮੂਹ ਵਿੱਚ ਚੱਲਦੀਆਂ ਹਨ ਅਤੇ ਤੁਸੀਂ ਇੱਥੇ ਵੀ ਚਲਾ ਸਕਦੇ ਹੋ।
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